SOLUTION: given that log‘a’(b^2)=c, express log‘a’(7b^5)+log‘a’(a^3)-log‘a’(7) in terms of c only. is the ‘a’ is in single quotation marks, it means the base of the lo

Question 1158662: given that log‘a’(b^2)=c, express log‘a’(7b^5)+log‘a’(a^3)-log‘a’(7) in terms of c only.
is the ‘a’ is in single quotation marks, it means the base of the log.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Throughout the entire solution below, all logs are base 'a'.
Instead of writing log'a'(x) to mean "log base 'a' of x", I'm simply going to write "log(x)".
Alternatively I could write log(a,x) to mean "log base 'a' of x", but again I'll just stick with log(x) for simplicity.

log(b^2) = c is given

We'll use the rule y*log(x) = log(x^y) to transform the exponent of '2' into an exponent of '5'

Multiply both sides by (5/2) and apply the log rule shown above
log(b^2) = c
(5/2)log(b^2) = (5/2)c
log( (b^2)^(5/2) ) = (5/2)c
log( b^(2*(5/2)) ) = (5/2)c
log( b^5 ) = (5/2)c
The exponents 2 and 5/2 multiply to 5.

We are able to replace log( b^5 ) with (5/2)c.

Next use the log rule log(xy) = log(x)+log(y)
log(7b^5) = log(7)+log(b^5)
log(7b^5) = log(7)+(5/2)c

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Therefore,
log(7b^5) + log(a^3) - log(7) = log(7)+(5/2)c + 3*log(a) - log(7)
log(7b^5) + log(a^3) - log(7) = log(7)+(5/2)c + 3*1 - log(7)
log(7b^5) + log(a^3) - log(7) = (5/2)c + 3 + log(7)-log(7)
log(7b^5) + log(a^3) - log(7) = (5/2)c + 3

Again, all logs in this solution are base 'a'.

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